3.200 \(\int \frac{1}{(a+b \cos ^{-1}(c x))^{5/2}} \, dx\)

Optimal. Leaf size=163 \[ -\frac{4 \sqrt{2 \pi } \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{3 b^{5/2} c}+\frac{4 \sqrt{2 \pi } \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{3 b^{5/2} c}+\frac{4 x}{3 b^2 \sqrt{a+b \cos ^{-1}(c x)}}+\frac{2 \sqrt{1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}} \]

[Out]

(2*Sqrt[1 - c^2*x^2])/(3*b*c*(a + b*ArcCos[c*x])^(3/2)) + (4*x)/(3*b^2*Sqrt[a + b*ArcCos[c*x]]) + (4*Sqrt[2*Pi
]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]])/(3*b^(5/2)*c) - (4*Sqrt[2*Pi]*FresnelC[(Sqr
t[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[a/b])/(3*b^(5/2)*c)

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Rubi [A]  time = 0.258632, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {4622, 4720, 4624, 3306, 3305, 3351, 3304, 3352} \[ -\frac{4 \sqrt{2 \pi } \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{3 b^{5/2} c}+\frac{4 \sqrt{2 \pi } \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{3 b^{5/2} c}+\frac{4 x}{3 b^2 \sqrt{a+b \cos ^{-1}(c x)}}+\frac{2 \sqrt{1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[c*x])^(-5/2),x]

[Out]

(2*Sqrt[1 - c^2*x^2])/(3*b*c*(a + b*ArcCos[c*x])^(3/2)) + (4*x)/(3*b^2*Sqrt[a + b*ArcCos[c*x]]) + (4*Sqrt[2*Pi
]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]])/(3*b^(5/2)*c) - (4*Sqrt[2*Pi]*FresnelC[(Sqr
t[2/Pi]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[b]]*Sin[a/b])/(3*b^(5/2)*c)

Rule 4622

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x])^(n + 1)
)/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcCos[c*x])^(n + 1))/Sqrt[1 - c^2*x^2], x], x] /; Fre
eQ[{a, b, c}, x] && LtQ[n, -1]

Rule 4720

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp
[((f*x)^m*(a + b*ArcCos[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] + Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)
^(m - 1)*(a + b*ArcCos[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && GtQ[d, 0]

Rule 4624

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/(b*c), Subst[Int[x^n*Sin[a/b - x/b], x], x, a
+ b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x]

Rule 3306

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d +
f*x]/Sqrt[c + d*x], x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c
, d, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(f*x^2)/d],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \cos ^{-1}(c x)\right )^{5/2}} \, dx &=\frac{2 \sqrt{1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}+\frac{(2 c) \int \frac{x}{\sqrt{1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^{3/2}} \, dx}{3 b}\\ &=\frac{2 \sqrt{1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}+\frac{4 x}{3 b^2 \sqrt{a+b \cos ^{-1}(c x)}}-\frac{4 \int \frac{1}{\sqrt{a+b \cos ^{-1}(c x)}} \, dx}{3 b^2}\\ &=\frac{2 \sqrt{1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}+\frac{4 x}{3 b^2 \sqrt{a+b \cos ^{-1}(c x)}}-\frac{4 \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}-\frac{x}{b}\right )}{\sqrt{x}} \, dx,x,a+b \cos ^{-1}(c x)\right )}{3 b^3 c}\\ &=\frac{2 \sqrt{1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}+\frac{4 x}{3 b^2 \sqrt{a+b \cos ^{-1}(c x)}}+\frac{\left (4 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{x}{b}\right )}{\sqrt{x}} \, dx,x,a+b \cos ^{-1}(c x)\right )}{3 b^3 c}-\frac{\left (4 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{x}{b}\right )}{\sqrt{x}} \, dx,x,a+b \cos ^{-1}(c x)\right )}{3 b^3 c}\\ &=\frac{2 \sqrt{1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}+\frac{4 x}{3 b^2 \sqrt{a+b \cos ^{-1}(c x)}}+\frac{\left (8 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{x^2}{b}\right ) \, dx,x,\sqrt{a+b \cos ^{-1}(c x)}\right )}{3 b^3 c}-\frac{\left (8 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{x^2}{b}\right ) \, dx,x,\sqrt{a+b \cos ^{-1}(c x)}\right )}{3 b^3 c}\\ &=\frac{2 \sqrt{1-c^2 x^2}}{3 b c \left (a+b \cos ^{-1}(c x)\right )^{3/2}}+\frac{4 x}{3 b^2 \sqrt{a+b \cos ^{-1}(c x)}}+\frac{4 \sqrt{2 \pi } \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{3 b^{5/2} c}-\frac{4 \sqrt{2 \pi } C\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right ) \sin \left (\frac{a}{b}\right )}{3 b^{5/2} c}\\ \end{align*}

Mathematica [C]  time = 1.62112, size = 194, normalized size = 1.19 \[ \frac{2 \left (e^{-\frac{i a}{b}} \left (a+b \cos ^{-1}(c x)\right ) \left (e^{\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}}-\sqrt{-\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )\right )-e^{-i \cos ^{-1}(c x)} \left (a+b \cos ^{-1}(c x)\right ) \left (-1+e^{\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}} \sqrt{\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )\right )+b \sqrt{1-c^2 x^2}\right )}{3 b^2 c \left (a+b \cos ^{-1}(c x)\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCos[c*x])^(-5/2),x]

[Out]

(2*(b*Sqrt[1 - c^2*x^2] + ((a + b*ArcCos[c*x])*(E^((I*(a + b*ArcCos[c*x]))/b) - Sqrt[((-I)*(a + b*ArcCos[c*x])
)/b]*Gamma[1/2, ((-I)*(a + b*ArcCos[c*x]))/b]))/E^((I*a)/b) - ((a + b*ArcCos[c*x])*(-1 + E^((I*(a + b*ArcCos[c
*x]))/b)*Sqrt[(I*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, (I*(a + b*ArcCos[c*x]))/b]))/E^(I*ArcCos[c*x])))/(3*b^2*c*
(a + b*ArcCos[c*x])^(3/2))

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Maple [B]  time = 0.111, size = 324, normalized size = 2. \begin{align*}{\frac{2}{3\,{b}^{2}c} \left ( 2\,\arccos \left ( cx \right ) \sqrt{2}\sqrt{\pi }\sqrt{a+b\arccos \left ( cx \right ) }\sqrt{{b}^{-1}}\cos \left ({\frac{a}{b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) b-2\,\arccos \left ( cx \right ) \sqrt{2}\sqrt{\pi }\sqrt{a+b\arccos \left ( cx \right ) }\sqrt{{b}^{-1}}\sin \left ({\frac{a}{b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) b+2\,\sqrt{2}\sqrt{\pi }\sqrt{a+b\arccos \left ( cx \right ) }\sqrt{{b}^{-1}}\cos \left ({\frac{a}{b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) a-2\,\sqrt{2}\sqrt{\pi }\sqrt{a+b\arccos \left ( cx \right ) }\sqrt{{b}^{-1}}\sin \left ({\frac{a}{b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) a+2\,\arccos \left ( cx \right ) \cos \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ) b+\sin \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ) b+2\,\cos \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ) a \right ) \left ( a+b\arccos \left ( cx \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arccos(c*x))^(5/2),x)

[Out]

2/3/c/b^2*(2*arccos(c*x)*2^(1/2)*Pi^(1/2)*(a+b*arccos(c*x))^(1/2)*(1/b)^(1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/
2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*b-2*arccos(c*x)*2^(1/2)*Pi^(1/2)*(a+b*arccos(c*x))^(1/2)*(1/b)^(1/2)
*sin(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*b+2*2^(1/2)*Pi^(1/2)*(a+b*arccos(c*
x))^(1/2)*(1/b)^(1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*a-2*2^(1/2)*Pi
^(1/2)*(a+b*arccos(c*x))^(1/2)*(1/b)^(1/2)*sin(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arccos(c*x))^(1
/2)/b)*a+2*arccos(c*x)*cos((a+b*arccos(c*x))/b-a/b)*b+sin((a+b*arccos(c*x))/b-a/b)*b+2*cos((a+b*arccos(c*x))/b
-a/b)*a)/(a+b*arccos(c*x))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arccos(c*x))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arccos(c*x) + a)^(-5/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arccos(c*x))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \operatorname{acos}{\left (c x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*acos(c*x))**(5/2),x)

[Out]

Integral((a + b*acos(c*x))**(-5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arccos(c*x))^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccos(c*x) + a)^(-5/2), x)